Show that $|f(a+tu)-f(a)|\leq \max_{1\leq i\leq n}|f(a\pm te_i )- f(a)|$ where $f:\mathbb{R}^n\to \mathbb{R}$ is function convex.

Question

Show that $|f(a+tu)-f(a)|\leq \max_{1\leq i\leq n}|f(a\pm te_i )- f(a)|$ where $f:\mathbb{R}^n\to \mathbb{R}$ is function convex and $(e_1,e_2,\cdots ,e_n)$ is the canonical basis of $\mathbb{R}^n$ and $x,u,a\in \mathbb{R}^n$ with $||u||_1=\sum_{1}^{n}|u_i|=1.$ Also $t\in [0,1].$

I tried to create something of the form $f((1-t)x+ ty)$ but nothing seems to work. Maybe there is some other fact about convex functions that I need to use here. Perhaps someone can give a hint for this problem.

Answer 1

Try to use $a + t\ u = \sum_i^n |u_i| (a + sign(u_i)t\ e_i)$, i.e. $a+t\ u$ is a convex combination of $(a + sign(u_i)t\ e_i)$ ($i=1,..,n$).